Problem: $\dfrac{ 4n + 3p }{ -8 } = \dfrac{ 5n - 8q }{ -6 }$ Solve for $n$.
Answer: Multiply both sides by the left denominator. $\dfrac{ 4n + 3p }{ -{8} } = \dfrac{ 5n - 8q }{ -6 }$ $-{8} \cdot \dfrac{ 4n + 3p }{ -{8} } = -{8} \cdot \dfrac{ 5n - 8q }{ -6 }$ $4n + 3p = -{8} \cdot \dfrac { 5n - 8q }{ -6 }$ Multiply both sides by the right denominator. $4n + 3p = -8 \cdot \dfrac{ 5n - 8q }{ -{6} }$ $-{6} \cdot \left( 4n + 3p \right) = -{6} \cdot -8 \cdot \dfrac{ 5n - 8q }{ -{6} }$ $-{6} \cdot \left( 4n + 3p \right) = -8 \cdot \left( 5n - 8q \right)$ Distribute both sides $-{6} \cdot \left( 4n + 3p \right) = -{8} \cdot \left( 5n - 8q \right)$ $-{24}n - {18}p = -{40}n + {64}q$ Combine $n$ terms on the left. $-{24n} - 18p = -{40n} + 64q$ ${16n} - 18p = 64q$ Move the $p$ term to the right. $16n - {18p} = 64q$ $16n = 64q + {18p}$ Isolate $n$ by dividing both sides by its coefficient. ${16}n = 64q + 18p$ $n = \dfrac{ 64q + 18p }{ {16} }$ All of these terms are divisible by $2$ $n = \dfrac{ {32}q + {9}p }{ {8} }$